Question

A transformer is used to light a $100W$ and $110V$ lamp from $220V$ mains. If the main current is $0.5$ amp, the efficiency of the transformer is approximately:

• (A) $50\mathrm{\%}$
• (B) $90\mathrm{\%}$
• (C) $10\mathrm{\%}$
• (D) $30\mathrm{\%}$

Answer 1

The Correct Option is B

Explanation:
Given,Output power $P=100W$Voltage across primary $V_p=220V$Current in the primary $I_p=0.5A$Efficiency of a transformer$\eta =\frac{\text{ output power }}{\text{ input power }}\times 100$$=\frac{P}{V_p{I}_p}\times 100$$=\frac{100}{220\times 0.5}\times 100=90\mathrm{\%}$Hence, the correct option is (B).

Answer 2

Given:
- Power of the lamp: \( P_{\text{lamp}} = 100 \text{ W} \)
- Voltage of the lamp: \( V_{\text{lamp}} = 110 \text{ V} \)
- Voltage of the mains: \( V_{\text{mains}} = 220 \text{ V} \)
- Main current: \( I_{\text{mains}} = 0.5 \text{ A} \)
 Steps to solve:
1. Calculate the power input to the transformer:
 The power input (\(P_{\text{in}}\)) is given by the product of the mains voltage and the mains current:
  \[ P_{\text{in}} = V_{\text{mains}} \times I_{\text{mains}} \]
  \[ P_{\text{in}} = 220 \text{ V} \times 0.5 \text{ A} \]
  \[ P_{\text{in}} = 110 \text{ W} \]
2. Power output of the transformer:
 The power output (\(P_{\text{out}}\)) is the power consumed by the lamp:
  \[ P_{\text{out}} = P_{\text{lamp}} = 100 \text{ W} \]
3. Calculate the efficiency of the transformer:
The efficiency (\(\eta\)) of the transformer is given by the ratio of the output power to the input power, expressed as a percentage:
  \[ \eta = \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) \times 100\% \]
  \[ \eta = \left( \frac{100 \text{ W}}{110 \text{ W}} \right) \times 100\% \]
  \[ \eta \approx 90.91\% \]
Final Result:
The efficiency of the transformer is approximately:
\[ \boxed{90.91\%} \]